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I recently have been reading Basic Mathematics by Serge Lang and encountered this question

Let $z$ be a complex number not equal to $0$. Let $n$ be a positive integer. Show that there are $n$ distinct complex numbers $w$ such that $w^n = z$. Write these complex numbers in polar form. The proof given that a polynomial of degree $n$ has at most $n$ roots applies to the complex case, and thus we see that there are no other complex numbers $w$ such that $w^n = z$ other than those you have presumably written down.

I tried relating it with polynomial in form $x^n -z = 0$, and proceeding with that same pattern and tried to come up with solutions, but it gets quite complicated pretty quick and I wonder if it will worth to consider odd/even powers and their effect on imaginary part.

I would also love to see if there is any way to prove the $n$ roots thing for complex number case.

I will appreciate the help. Thanks!

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The fundamental theorem of algebra guarantees the $n$ solutions.

Take a primitive $n$-th root of unity: $\zeta=e^{2\pi i/n}$. Then the $n$ solutions are $\zeta^kz^{1/n},\,k=0,2,\dots n-1$.

If $z=re^{i\theta}$, we get $r^{1/n}e^{i(\theta+2\pi k)/n}$.

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  • $\begingroup$ Apparently, none of those are really mentioned in this text so far. Maybe fundamental theorem have been indirectly, but not root of unity. Is there any more "elementary" way to approach this? $\endgroup$
    – user773387
    Apr 15 '20 at 10:02
  • $\begingroup$ Not really. I have added a little concerning polar. $\endgroup$
    – user403337
    Apr 15 '20 at 15:51

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